A) 6.01
B) 6.04
C) 6.02
D) None of these
Correct Answer: A
Solution :
\[{{(217)}^{1/3}}={{({{6}^{3}}+1)}^{1/3}}=6{{\left( 1+\frac{1}{{{6}^{3}}} \right)}^{1/3}}\] On expansion by binomial theorem \[=6\,\,\left( 1+\frac{1}{3\times 216}-\frac{1\times 2}{3\times 3\times 2}{{\left( \frac{1}{216} \right)}^{2}}+..... \right)=6.01\]
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